Explore Westonci.ca, the leading Q&A site where experts provide accurate and helpful answers to all your questions. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
we know that θ is in the III Quadrant, keeping in mind that cosine as well as sine are both negative on that Quadrant.
we also know that sin(θ) = - 8/17, keeping in mind that the hypotenuse is just a radius unit and thus is never negative.
[tex]sin(\theta )=\cfrac{\stackrel{opposite}{-8}}{\underset{hypotenuse}{17}}\qquad \qquad \textit{let's find the \underline{adjacent} side} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{17^2-(-8)^2}=a\implies \pm\sqrt{225}=a\implies \pm 15=a\implies \stackrel{III~Quadrant}{-15=a} \\\\[-0.35em] ~\dotfill[/tex]
[tex]sec(\theta )=\cfrac{\stackrel{hypotenuse}{17}}{\underset{adjacent}{-15}}~\hspace{10em} cot(\theta )=\cfrac{\stackrel{adjacent}{-15}}{\underset{opposite}{-8}}\implies cot(\theta )=\cfrac{15}{8}[/tex]
Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.
